| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2002 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Code/password formation |
| Difficulty | Moderate -0.3 This is a straightforward application of combinations with clear structure: (i) is a simple verification of C(8,2)=28, (ii) sums four basic combination calculations, and (iii) applies the multiplication principle. The question requires only standard counting techniques with no problem-solving insight or tricky cases, making it slightly easier than average but still requiring proper understanding of combinations. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Part (i): \(4C_2 = 28\) or \(7+6+5+4+3+2+1\) | B1 | 1 |
| Part (ii): \(4C_1 + 4C_2 + 4C_3 + 4C_4 = 8 + 28 + 56 + 70\) | M1 | For listing 4 Combination options (can be added or multiplied here) |
| A1 | For \(4C_1 + 4C_2 + 4C_3 + 4C_4\) | |
| A1 | For at least 3 correct numbers, can be implied by seeing 878080 (mult) | |
| \(= 162\) | A1 | 4 |
| SR \(4C_1+4C_2+...+4C_4\) M1 only | ||
| SR \(4C_3 \times C_3 \times C_1 \times 4C_2\) M1 only | ||
| Part (iii): \((162)^4 = 688 \, 747 \, 536\) or \(3s\) | M1 | For \((their (ii))^4\) or \(4C_3+4C_4+4C_1 \times 4C_2\) |
| A1 ft | 2 | For correct answer in any form |
**Part (i):** $4C_2 = 28$ or $7+6+5+4+3+2+1$ | B1 | 1 | For $4C_2$
**Part (ii):** $4C_1 + 4C_2 + 4C_3 + 4C_4 = 8 + 28 + 56 + 70$ | M1 | For listing 4 Combination options (can be added or multiplied here)
| A1 | For $4C_1 + 4C_2 + 4C_3 + 4C_4$
| A1 | For at least 3 correct numbers, can be implied by seeing 878080 (mult)
$= 162$ | A1 | 4 | For correct answer
| | SR $4C_1+4C_2+...+4C_4$ M1 only
| | SR $4C_3 \times C_3 \times C_1 \times 4C_2$ M1 only
**Part (iii):** $(162)^4 = 688 \, 747 \, 536$ or $3s$ | M1 | For $(their (ii))^4$ or $4C_3+4C_4+4C_1 \times 4C_2$
| A1 ft | 2 | For correct answer in any form
---4 In a certain hotel, the lock on the door to each room can be opened by inserting a key card. The key card can be inserted only one way round. The card has a pattern of holes punched in it. The card has 4 columns, and each column can have either 1 hole, 2 holes, 3 holes or 4 holes punched in it. Each column has 8 different positions for the holes. The diagram illustrates one particular key card with 3 holes punched in the first column, 3 in the second, 1 in the third and 2 in the fourth.\\
\includegraphics[max width=\textwidth, alt={}, center]{2bcbd4d3-0d41-48fa-8f70-192b158c0bbe-2_410_214_1811_968}\\
(i) Show that the number of different ways in which a column could have exactly 2 holes is 28 .\\
(ii) Find how many different patterns of holes can be punched in a column.\\
(iii) How many different possible key cards are there?
\hfill \mbox{\textit{CAIE S1 2002 Q4 [7]}}